# Frequency of wave related with the speed of wave

Today I am studying about waves. So, we learn the formula "speed = frequency * wavelength". There is a question come out in my book

If the frequency of water waves is decreased, what will happened to its speed?

So, this is what did,

• If speed = frequency * wavelength
• That means speed is directly proportional to frequency
• When frequency increases, speed increases and vice versa.

Eventually, I write the speed of the water waves will decrease. But the answer from the book is the speed of water waves will stay constant.

So, my question is why the speed of water waves will stay constant if the frequency of water waves is decreased?

## 2 Answers

I will delay the use of water waves, rather use waves on a string, as in case of water waves, nonlinear effects can be there which can have different dispersion relations like in the case of shallow water/ deep water waves and thereby a different relation for group velocity than you mentioned. So I will derive the case for waves on a string and then sketch the analogous case for water waves, taking into account appropriate approximations. So for waves on a string we have

$$c = \nu \lambda$$

Now we know that the speed of the waves is dependent on the tension and the linear mass density.

$$c = \sqrt{T/\mu}$$ Since neither of them are changing in your example, we know that the speed of the waves is constant. Therefore from the first relation the only thing that changes with change with frequency is the wavelength and not speed.

For water waves, similarly, using appropriate limits if the dispersion relation yields a similar group velocity relation with frequency and wavelength, we can say that since the pressure and the density are not changed, and as a result, the only thing that changes with change with frequency is the wavelength and not speed.

The relation $c = \lambda f$ is derived from the propagation equation : $$\frac{1}{c^2} \frac{\partial^2u}{\partial t^2} = \frac{\partial^2u}{\partial x^2}$$

Where $c$ is in a fact a constant, by mathematical nature (it is defined from physical constants, not changing during propagation). So when stating that $c = \lambda f$, it is implicit that $c$ is a constant, which is not very clear in the lesson/book/whatever you are reading.

Therefore, your question's answer is : the wavelength $\lambda$ will increase to compensate for $f$ decrease.

And that's why in mechanical propagation as well as radiowaves, lower frequencies mean higher wavelengths.

• Surface waves on liquids often have dispersion (i.e. $c = c(f)$ rather than being consistent at all frequencies), so while $c(f)$ is constant it isn't kosher to say that "$c$ is constant" when comparing different frequencies. The effect is pretty small, so it is generally neglected in the first treatment. – dmckee --- ex-moderator kitten Mar 3 '16 at 15:01
• I tried to say it that way : "it is defined from physical constants, not changing during propagation" which does not target dispersion specifically, but any non-linear or lossy phenomenon – MaximGi Mar 3 '16 at 15:37